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Difference between revisions of "1996 AIME Problems/Problem 15"

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(solutions, (2) by Altheman)
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== Problem ==
 
== Problem ==
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In parallelogram <math>ABCD</math>, let <math>O</math> be the intersection of diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math>. Angles <math>CAB</math> and <math>DBC</math> are each twice as large as angle <math>DBA</math>, and angle <math>ACB</math> is <math>r</math> times as large as angle <math>AOB</math>. Find the greatest integer that does not exceed <math>1000r</math>.
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In [[parallelogram]] <math>ABCD</math>, let <math>O</math> be the intersection of [[diagonal]]s <math>\overline{AC}</math> and <math>\overline{BD}</math>. Angles <math>CAB</math> and <math>DBC</math> are each twice as large as angle <math>DBA</math>, and angle <math>ACB</math> is <math>r</math> times as large as angle <math>AOB</math>. Find the [[floor function|greatest integer]] that does not exceed <math>1000r</math>.
  
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__TOC__
 
== Solution ==
 
== Solution ==
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{{solution}}
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=== Solution 1 ===
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<center><asy>
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</asy></center>
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Let <math>\theta = \angle DBA</math>. Then <math>\angle CAB = \angle DBD = 2\theta</math>, <math>\angle AOB = 180 - 3\theta</math>, and <math>\angle ACB = 180 - 5\theta</math>. Since <math>ABCD</math> is a parallelogram, it follows that <math>OA = OC</math>. By the [[Law of Sines]] on <math>\triangle ABO,\, \triangle BCO</math>,
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<center><math>\frac{\sin \angle CBO}{OC} = \frac{\sin \angle ACB}{OB} \quad \text{and} \quad \frac{\sin \angle DBA}{OC} = \frac{\sin \angle BAC}{OB}.</math></center>
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Dividing the two equalities yields
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<cmath>\frac{\sin 2\theta}{\sin \theta} = \frac{\sin (180 - 5\theta)}{\sin 2\theta} \Longrightarrow \sin^2 2\theta = \sin 5\theta \sin \theta.</cmath>
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Pythagorean and product-to-sum identities yield
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<cmath>1 - \cos^2 2 \theta = \cos 4\theta - \cos 6 \theta,</cmath>
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and the double and triple angle (<math>\cos 3x = 4\cos^3 x - 3\cos x</math>) formulas further simplify this to
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<cmath>4\cos^3 2\theta - 4\cos^2 2\theta - 3\cos \theta + 3 = (4\cos^2 2\theta - 3)(\cos 2\theta - 1) = 0</cmath>
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The only value of <math>\theta</math> that fits in this context comes from <math>4 \cos^2 2\theta - 3 = 0 \Longrightarrow \cos 2\theta = \frac{\sqrt{3}}{2} \Longrightarrow \theta = 15^{\circ}</math>. The answer is <math>\lfloor 1000r \rfloor = \left\lfloor 1000 \cdot \frac{180 - 5\theta}{180 - 3\theta} \right\rfloor = \left \lfloor \frac{7000}{9} \right \rfloor = \boxed{777}</math>.
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=== Solution 2 === <!-- I would put this solution first, but needs a bit of formatting for clarity - azjps -->
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We will focus on <math>\triangle ABC</math>. Let <math>\angle ABO = x</math>, so <math>\angle BAO = \angle OBC = 2x</math>. Draw the [[perpendicular]] from <math>C</math> intersecting <math>AB</math> at <math>H</math>. [[Without loss of generality]], let <math>AO = CO = 1</math>. Then <math>HO = 1</math>, since <math>O</math> is the [[circumcenter]] of <math>\triangle AHC</math>. Then <math>\angle OHA = 2x</math>.
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By the Exterior Angle Theorem, <math>\angle COB = 3x</math> and <math>\angle COH = 4x</math>. That implies that <math>\angle HOB = x</math>. That makes <math>HO = HB = 1</math>. Then since by AA (<math>\angle HBC = \angle HOC = 3x</math> and reflexive on <math>\angle OCB</math>), <math>\triangle OCB \sim \triangle BCA</math>.
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<center><math>\frac {CO}{BC} = \frac {BC}{AC} \implies 2 = BC^2 = \implies BC = \sqrt {2}. </math></center>
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Then by the [[Pythagorean Theorem]], <math>1^2 + HC^2 = \left(\sqrt {2}\right)^2\implies HC = 1</math>. That makes <math>\triangle HOC</math> equilateral. Then <math>\angle HOC = 4x = 60 \implies x = 15</math>. Then <math>\angle AOB = 180 - 3 \times 15 = 135^{\circ}</math> and <math>\angle ACB = 180 - 5 \times 15 = 105^{\circ}</math>.
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Then <math>\frac {\angle ACB}{\angle AOB} = r = \frac {7}{9}</math>. Then it follows that <math>\left\lfloor \frac {7000}{9}\right\rfloor = \boxed{777}</math>.
  
 
== See also ==
 
== See also ==
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*[[1996 AIME Problems]]
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{{AIME box|year=1996|num-b=14|after=Final Problem}}
  
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{{AIME box|year=1996|num-b=14|after=Final Problem}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 21:53, 13 August 2008

Problem

In parallelogram $ABCD$, let $O$ be the intersection of diagonals $\overline{AC}$ and $\overline{BD}$. Angles $CAB$ and $DBC$ are each twice as large as angle $DBA$, and angle $ACB$ is $r$ times as large as angle $AOB$. Find the greatest integer that does not exceed $1000r$.

Solution

Solution 1

[asy] [/asy]

Let $\theta = \angle DBA$. Then $\angle CAB = \angle DBD = 2\theta$, $\angle AOB = 180 - 3\theta$, and $\angle ACB = 180 - 5\theta$. Since $ABCD$ is a parallelogram, it follows that $OA = OC$. By the Law of Sines on $\triangle ABO,\, \triangle BCO$,

$\frac{\sin \angle CBO}{OC} = \frac{\sin \angle ACB}{OB} \quad \text{and} \quad \frac{\sin \angle DBA}{OC} = \frac{\sin \angle BAC}{OB}.$

Dividing the two equalities yields

\[\frac{\sin 2\theta}{\sin \theta} = \frac{\sin (180 - 5\theta)}{\sin 2\theta} \Longrightarrow \sin^2 2\theta = \sin 5\theta \sin \theta.\]

Pythagorean and product-to-sum identities yield

\[1 - \cos^2 2 \theta = \cos 4\theta - \cos 6 \theta,\]

and the double and triple angle ($\cos 3x = 4\cos^3 x - 3\cos x$) formulas further simplify this to

\[4\cos^3 2\theta - 4\cos^2 2\theta - 3\cos \theta + 3 = (4\cos^2 2\theta - 3)(\cos 2\theta - 1) = 0\]

The only value of $\theta$ that fits in this context comes from $4 \cos^2 2\theta - 3 = 0 \Longrightarrow \cos 2\theta = \frac{\sqrt{3}}{2} \Longrightarrow \theta = 15^{\circ}$. The answer is $\lfloor 1000r \rfloor = \left\lfloor 1000 \cdot \frac{180 - 5\theta}{180 - 3\theta} \right\rfloor = \left \lfloor \frac{7000}{9} \right \rfloor = \boxed{777}$.

Solution 2

We will focus on $\triangle ABC$. Let $\angle ABO = x$, so $\angle BAO = \angle OBC = 2x$. Draw the perpendicular from $C$ intersecting $AB$ at $H$. Without loss of generality, let $AO = CO = 1$. Then $HO = 1$, since $O$ is the circumcenter of $\triangle AHC$. Then $\angle OHA = 2x$.

By the Exterior Angle Theorem, $\angle COB = 3x$ and $\angle COH = 4x$. That implies that $\angle HOB = x$. That makes $HO = HB = 1$. Then since by AA ($\angle HBC = \angle HOC = 3x$ and reflexive on $\angle OCB$), $\triangle OCB \sim \triangle BCA$.

$\frac {CO}{BC} = \frac {BC}{AC} \implies 2 = BC^2 = \implies BC = \sqrt {2}.$

Then by the Pythagorean Theorem, $1^2 + HC^2 = \left(\sqrt {2}\right)^2\implies HC = 1$. That makes $\triangle HOC$ equilateral. Then $\angle HOC = 4x = 60 \implies x = 15$. Then $\angle AOB = 180 - 3 \times 15 = 135^{\circ}$ and $\angle ACB = 180 - 5 \times 15 = 105^{\circ}$.

Then $\frac {\angle ACB}{\angle AOB} = r = \frac {7}{9}$. Then it follows that $\left\lfloor \frac {7000}{9}\right\rfloor = \boxed{777}$.

See also

1996 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Final Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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